#refractive index

Optical Phenomena and Applications

What is the wavelength of light in a medium with refractive index 1.5 if its wavelength in vacuum is \( 750 \, \text{nm}

**Polarization requires transverse waves** because only transverse can have orientation perpendicular to propagation, longitudinal cannot be polarized, wave theory requires light transverse to explain polarization, polaroids transmit only component along pass-axis, unpolarized has random transverse orientations, after polaroid polarized. Wavelength in a medium λ_m = (λvₐcuuₘ/n) . Given λvₐcuuₘ = 750 nm , n = 1.5 , λ_m = (750/1.5) = 500 nm . Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I = I₀ cos²θ, n = c/v, sinC = 1/n, λ' = λ/n and A = 2a cos(φ/2), calculation gives 500 nm, illustrating interf

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Optical Phenomena and Applications

What is the refractive index of a medium if the speed of light in it is \( 1.8 \times 10^8 \, \text{m/s} \) and in vacuu

**Incoherent sources** intensity adds I = I₁+I₂, no interference pattern because phase random, two independent sources cannot produce stable interference because phase difference fluctuates rapidly, coherent sources required with constant phase, laser coherent, visibility of fringes requires coherence, degree of coherence determines contrast. Refractive index n = (c/v) . c = 3.0 × 10⁸ m/s , v = 1.8 × 10⁸ m/s . n = (3.0 × 10⁸/1.8 × 10⁸) ≈ 1.67 . Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I = I₀ cos²θ, n = c/v, sinC = 1/n, λ' = λ/n and A = 2a

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Optical Phenomena and Applications

What is the critical angle for light passing from a medium with refractive index 1.85 to air (refractive index 1.0)?

**Polarization requires transverse waves** because only transverse can have orientation perpendicular to propagation, longitudinal cannot be polarized, wave theory requires light transverse to explain polarization, polaroids transmit only component along pass-axis, unpolarized has random transverse orientations, after polaroid polarized. sin i_c = (n₂/n₁) , where n₁ = 1.85 , n₂ = 1.0 . sin i_c = (1.0/1.85) ≈ 0.541 , i_c = sin⁻¹(0.541) ≈ 32.7° . Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I = I₀ cos²θ, n = c/v, sinC = 1/n, λ' = λ/n and A = 2a cos(φ/2), calculation gives 32.7°, illustrating int

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Optical Phenomena and Applications

What is the refractive index of a medium if the critical angle for light passing into air is \( 39^\circ \)?

**Convex lens focusing** plane wave into point because lens introduces phase delay proportional to thickness, converting plane wavefront to spherical converging to focal point, property ensures rays parallel to axis meet at focus, spherical aberration minimized for paraxial rays, lensmaker's formula determines focal length. sin i_c = (n₂/n₁) , where n₂ = 1.0 (air), i_c = 39° . sin 39° ≈ 0.629 , n₁ = (1.0/0.629) ≈ 1.59 . Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I = I₀ cos²θ, n = c/v, sinC = 1/n, λ' = λ/n and A = 2a cos(φ/2), calculation gives 1.59,

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Optical Phenomena and Applications

What is the speed of light in a medium if its refractive index is 1.2 and the speed in vacuum is \( 3.0 \times 10^8 \, \

**Polarization requires transverse waves** because only transverse can have orientation perpendicular to propagation, longitudinal cannot be polarized, wave theory requires light transverse to explain polarization, polaroids transmit only component along pass-axis, unpolarized has random transverse orientations, after polaroid polarized. Speed in a medium v = (c/n) . Given n = 1.2 , c = 3.0 × 10⁸ m/s , v = (3.0 × 10⁸/1.2) = 2.5 × 10⁸ m/s . Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I = I₀ cos²θ, n = c/v, sinC = 1/n, λ' = λ/n and A = 2a cos(φ/2), calculation gives

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Wave Properties, Frequency and Energy Conservation

What is the speed of light in a medium with refractive index 1.25, given the speed in vacuum is \( 3.0 \times 10^8 \, \t

**Frequency of light** remains unchanged when refracts from air into water because frequency determined by source, energy E= h f conserved, speed decreases v= c/n, wavelength decreases λ'=v/f= λ/n, energy of wave proportional to amplitude² not speed, intensity I =½ c ε₀ E₀², energy not depend on speed directly, when speed decreases amplitude may change but energy conserved, interference redistributes energy, total energy same, bright regions gain from dark. Speed in a medium v = (c/n) . Given n = 1.25 , c = 3.0 × 10⁸ m/s , v = (3.0 × 10⁸/1.25) = 2.4 × 10⁸ m/s . Using Δ = d sinθ,

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Wave Properties, Frequency and Energy Conservation

What is the wavelength of light in a medium with refractive index 1.25 if its wavelength in air is \( 625 \, \text{nm} \

**Frequency of light** remains unchanged when refracts from air into water because frequency determined by source, energy E= h f conserved, speed decreases v= c/n, wavelength decreases λ'=v/f= λ/n, energy of wave proportional to amplitude² not speed, intensity I =½ c ε₀ E₀², energy not depend on speed directly, when speed decreases amplitude may change but energy conserved, interference redistributes energy, total energy same, bright regions gain from dark. Wavelength in a medium λ_m = (λₐir/n) . Given λₐir = 625 nm , n = 1.25 , λ_m = (625/1.25) = 500 nm . Using Δ = d sinθ, y = n λ D/d, a

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Refraction, Refractive Index and Critical Angle

What is the speed of light in a medium with refractive index 1.8, given the speed in vacuum is \( 3.0 \times 10^8 \, \te

**Frequency of light** does not change on refraction, c = f λ, when enters denser medium speed decreases, wavelength decreases λ' = v/f = c/(n f) = λ/n, frequency same, for λ=480 nm f=c/λ=3×10⁸/480×10⁻⁹=6.25×10¹⁴ Hz, for 630 nm f=4.76×10¹⁴ Hz, for 540 nm f=5.56×10¹⁴ Hz, refractive index determines bending. Speed in a medium v = (c/n) . Given n = 1.8 , c = 3.0 × 10⁸ m/s , v = (3.0 × 10⁸/1.8) ≈ 1.67 × 10⁸ m/s . Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I = I₀ cos²θ, n = c/v, sinC = 1/n,

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Refraction, Refractive Index and Critical Angle

What is the refractive index of a medium if the speed of light in it is \( 2.5 \times 10^8 \, \text{m/s} \) and in vacuu

**Frequency of light** does not change on refraction, c = f λ, when enters denser medium speed decreases, wavelength decreases λ' = v/f = c/(n f) = λ/n, frequency same, for λ=480 nm f=c/λ=3×10⁸/480×10⁻⁹=6.25×10¹⁴ Hz, for 630 nm f=4.76×10¹⁴ Hz, for 540 nm f=5.56×10¹⁴ Hz, refractive index determines bending. Refractive index n = (c/v) . c = 3.0 × 10⁸ m/s , v = 2.5 × 10⁸ m/s . n = (3.0 × 10⁸/2.5 × 10⁸) = 1.2 . Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I = I₀ cos²θ, n = c/v, sinC = 1/n, λ'

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Refraction, Refractive Index and Critical Angle

What is the speed of light in glass if its refractive index is 1.6 and the speed of light in vacuum is \( 3.0 \times 10^

**Refractive index** n = c/v, c=3×10⁸ m/s vacuum, v speed in medium, for v=2.25×10⁸ m/s n=3/2.25=1.33, for v=1.8×10⁸ n=1.67, for n=1.3 v=3×10⁸/1.3=2.31×10⁸ m/s, for n=1.6 v=1.875×10⁸ m/s, for n=1.2 λ_medium = λ_vacuum/n =600/1.2=500 nm, for n=1.75 λ=700/1.75=400 nm, for n=1.5 λ=750/1.5=500 nm, wavelength in medium λ' = λ/n. Speed in a medium v = (c/n) . Given n = 1.6 , c = 3.0 × 10⁸ m/s , v = (3.0 × 10⁸/1.6) = 1.875 × 10⁸ m/s . Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I = I₀ cos²θ, n = c/v, sinC = 1/n, λ'

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Refraction, Refractive Index and Critical Angle

What is the critical angle for light passing from a medium with refractive index 1.9 to air (refractive index 1.0)?

**Critical angle** sinC = n₂/n₁, n₁ denser, n₂ rarer 1 for air, for C=36.9° n₁=1/sin36.9°=1/0.6=1.67, for C=39° n=1/sin39°=1/0.629=1.59, for n=1.85 C=arcsin(1/1.85)=arcsin0.5405=32.7°, total internal reflection occurs only when light travels from denser to rarer and incidence > C, because Snell's law would require sinθ₂>1 impossible. sin i_c = (n₂/n₁) , where n₁ = 1.9 , n₂ = 1.0 . sin i_c = (1.0/1.9) ≈ 0.526 , i_c = sin⁻¹(0.526) ≈ 31.8° . Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I = I₀ cos²θ, n = c/v, sinC = 1/n, λ' = λ/n and A = 2a cos(φ/2), calculation

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Refraction, Refractive Index and Critical Angle

What is the speed of light in a medium with a refractive index of 1.5, given the speed of light in vacuum is \( 3.0 \tim

**Frequency of light** does not change on refraction, c = f λ, when enters denser medium speed decreases, wavelength decreases λ' = v/f = c/(n f) = λ/n, frequency same, for λ=480 nm f=c/λ=3×10⁸/480×10⁻⁹=6.25×10¹⁴ Hz, for 630 nm f=4.76×10¹⁴ Hz, for 540 nm f=5.56×10¹⁴ Hz, refractive index determines bending. Refractive index n = (c/v) , where c is the speed in vacuum and v is the speed in the medium. Given n = 1.5 , v = (c/n) = (3.0 × 10⁸/1.5) = 2.0 × 10⁸ m/s . Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I =

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